ASRJC Alternating Current

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9749 18-1 Additional Notes Topic 18: Alternating Current Content: A Characteristics of alternating currents B Rectification with a diode C The transformer Learning Outcomes: Candidates should be able to: (a) show an understanding of and use the terms period, frequency, peak value and root-mean-square (r.m.s.) value as applied to an alternating current or voltage (b) deduce that the mean power in a resistive load is half the maximum (peak) power for a sinusoidal alternating current (c) represent an alternating current or an alternating voltage by an equation of the form x = x0 sin ωt (d) distinguish between r.m.s. and peak values and recall and solve problems using the relationship 0 /2rmsII = for the sinusoidal case (e) show an understanding of the principle of o peration of a simple iron -cored transformer and recall and solve problems using / / /s p s p p sN N V V I I== for an ideal transformer (f) explain the use of a single diode for the half-wave rectification of an alternating current. A Characteristics of Alternating Currents Understanding the Nature of Scientific Knowledge During the ‘War of the Currents” in the late 1880s, Thomas Edison (1847 -1931) favoured direct current while George Westinghouse (1846-1914) argued for alternating current. Direct current at 110V used to be the standard for electricity distribution in United States in terms of safety and practicality. However, the use of alternating current eventually prevailed as it was more effective in minimising power loss with the use of transformers and high voltage in long -distance transmission. Historical competition between companies for electrical power distributio n and other factors resulted in different standards for the mains voltage in different countries. Having different voltages and frequencies between countries, and for some, within the country, can create confusion and result in extra costs for electrical appliances and adaptors. In Singapore, 230V at 50 Hz is used. In line with technological changes in the way the society uses electricity, there is ongoing research to continue to find ways to minimize power losses in electrical power transmission. It appear s that direct current now has an opportunity to dominate many mo dern electrical digital devices . The method that will eventually triumph in the future will be the one that results in the least power losses.

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9749 18-2 Additional Notes A.1 Introduction to Alternating Currents An Alternating Current is a current that varies periodically in direction. Direct Current Alternating Current Polarity of d.c. supply remains constant with time. Polarity of a.c. supply changes with time. Current is unidirectional, from positive to negative terminal. Current changes direction. Current may increase and decrease in magnitude. Current may increase and decrease in magnitude. • The effects of a.c. are essentially the same as those of d.c. Both can be used for heating and lighting purposes. Check Your Understanding 1 Which of the following are examples of direct current and which are alternating current? A.2 Root-Mean-Square Value of an Alternating Current / EMF The root-mean-square value of an alternating current is the value of the steady direct current which produces heat at the same rate as the alternating current in a given resistor. time current time current time current time current time current A B C D E The root-mean-square voltage is defined as the steady direct voltage which produces heat at the same rate as the alternating voltage across a given resistor.

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9749 18-3 Additional Notes Worked Example Consider a periodic alternating current waveform Iac with period T of 8.0 s, passing through a resistor as shown below. The root-mean-square current Irms is mathematically determined by calculating the square root of the mean value of the square of the instantaneous current over one cycle. 2 2 2 0 area under curve for one cycle one period T rms ac dt T= = =  I III ( ) 222 4 3 4 2.55A8  + − ==rmsI The root-mean-square value is therefore the effective value of the alternating current that produces the same average power dissipation in a resistor as a steady direct current of the same value. This means that for the waveform above, the same average power is dissipated in the resistor if the alternating current is replaced by a steady direct current of 2.55 A. Note that this formula also applies to the r.m.s. value of an alternating e.m.f. source, as shown in Example 1. Example 1 The variation with time t of an alternating e.m.f. source E is as shown below. 0 1.0 2.0 3.0 -1.0 -2.0 -3.0 Iac / A t / s 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 0 1.0 2.0 3.0 -1.0 -2.0 -3.0 E / V t / s 2.0 4.0 6.0 8.0 10.0 12.0 14.0 16.0 Recall the concept of root- mean-square from Thermal Physics (root- mean-square speed of molecules). Steps to calculate r.m.s. value: 1. square 2. average 3. root

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9749 18-4 Additional Notes Determine, for the alternating e.m.f. source, the (a) Frequency [0.125 Hz] (b) Peak e.m.f. [3.0 V] (c) Root-mean-square e.m.f. [2.06 V] Solutions Thinking Process Parts (a) and (b) involve direct application of concepts and reading off the graph. Key question on objective: How can the root-mean-square e.m.f. be calculated from a graph? • The sequence should be: 1. Square the values of e.m.f. for one complete cycle of the a.c. 2. Average the squared values over one period 3. Square root the average Key questions to solve the problem: What is meant by ‘squaring the values’ when applied to a graph? • For this graph, ‘square’ refers to squaring the value of E at each point in time • Graphically, this can be seen by sketching a E2 against time graph. What is meant by ‘averaging over one period’ when applied here? • Sum up all the values of E2 multiplied by the time they are sustained over, then divide by the period • Graphically, this can be seen as dividing the area under the E2-t graph for one period by one period.

ANDERSON SERANGOON JUNIOR COLLEGE PHYSICS 9749 18-5 Additional Notes A.3 Power Dissipation by a Sinusoidal Alternating Current / EMF • The simplest alternating e.m.f. can be represented by the sinusoidal waveform. • Consider two circuits: the first with a d.c. source of E0, the second a sinusoidal a.c. source with peak value E0 and frequency f connected to a resistor in the diagram s below. • The table below illustrates the differences in current, e.m.f. and power dissipation for both circuits. Quantity DC Circuit AC Circuit Instantaneous EMF Edc = E0 Instantaneous, peak and mean e.m.f. are the same in value. Eac = E0 sin ω t Peak EMF E0 r.m.s. EMF 2 0 2 rms ac EEE == See mathematical proof on next page Instantaneous Current 0 dc EI R= Instantaneous, peak and mean current are the same in value. 00sin sinac ac E E t EIt R R R  = = =  Peak Current 0 0 EI R= r.m.s. Current 2 0 2 rms III == Instantaneous Power Dissipated in Resistor 2 2 0 dc dc EP I R R R ==  Instantaneous, peak and mean power are the same in value. 2 2 2 0 sinacP I R I R t == Peak Power Dissipated in Resistor 2 00P I R= Mean Power Dissipated in Resistor 00 0 0 0 22 22 rms rms IEP I E I E PP   ==      == Mean power = 1 2 Peak Power (only f